Posts tagged with: guesstimate

Hi,

I am struggling with a question in the guesstimate section. Although you did a good job of clarifying I am still a bit confused as math isn’t my strongest suit. The question is:
p. 52, Q.20: In the figure above A is the centre of the circle, A and D lie on BF and CE respectively, and B, D, F, and G lie on the circle. If BC=3, and DG bisects BF, what is the total area of the shaded regions?

I am just wondering, how you would go about calculating the area of the shaded region in detail.

PWN_the_SAT_Math_Guide__3rd_Edition__edited_3_15_15_pdf

OK, so here are the key things: BCEF is a rectangle, BF and DG are diameters, and AD = 3 just like BC does. So the circle has a radius of 3.

Triangle GEC has a base and a height of 6. Therefore, the smaller, similar triangle with a height of GA has a base and a height of 3.

The shaded region will be the area of half the circle minus the area of that small triangle. That’s \dfrac{\pi 3^2}{2}-\dfrac{1}{2}(3)(3) = \dfrac{9\pi - 9}{2}.

As anyone who’s ever chewed on a pencil knows, it doesn’t take much force to put a dent in a regular old #2 pencil. You might have an opportunity to use this to your advantage on the SAT. Occasionally, a geometry question will appear that asks you to figure out the length of a segment based on information about the lengths of other segments. If the figure is drawn to scale, you and your thumb nail are in business.

I don’t want to oversell this. You’re not going to see a huge score increase because of it. But if you desperately need to compare the lengths of a few segments, you can make notches in your pencil with your thumb nail, and then use your new makeshift ruler to solve the problem. Let’s look at an example where this might come in handy.

  1. In the figure above, PTACis the midpoint of perpendicular segments BD and PT, and D is the midpoint of AC. If AC = 4, BD = 8, and PQ = 3, what is the length of PT?(A) 6
    (B) 8
    (C) 10
    (D) 10.5
    (E) 11

So there’s a mathematical solution to this, of course, but since it’s drawn to scale and all we’re asked to do is compare the lengths of a couple line segments, I’m going to show you how to do this one the quick and dirty way. (Commenters, feel free to post the geometric solution if you like.)

Buckle up, things are about to get meta:

That’s right. I printed my own blog, and took pictures of it, and now I’m posting them on my blog. But seriously though, do you see what I’ve done here? I know BD = 8, and since the figure is drawn to scale, I can make thumbnail dents in my pencil and use them to approximate other lengths. Let’s look at PT:

OMG. It looks like PT = 8 too! In fact, there aren’t any other choices that are close enough to 8 for me to do any more work. I’m going to pick (B) and move right along.

Now, let me repeat what I said above: this trick is not going to win you beaucoup points; you might go through an entire test without an opportunity to use it. But it’s one more way to attack a geometry question that’s drawn to scale, and it just might help extricate you from a bind. Keep it in the back of your mind.

Here’s an important thing to remember: all figures on the SAT are drawn to scale unless indicated otherwise. In other words, if it doesn’t say “Note: figure not drawn to scale,” underneath it, it is drawn to scale. Most figures on the SAT are drawn to scale, which means it’s a good idea to guesstimate whenever possible.

Guesstimating could mean actively trying to eyeball relative angle measures, areas, or segment lengths, or it could mean sliding pieces of the diagram around in your mind. You might still end up doing some math because guesstimating doesn’t lead you all the way to an answer. But it’s important that you not waste the opportunity when a diagram is drawn to scale. Let’s dig right into an example:

  1. The figure above depicts two intersecting diameters of two concentric circles of radius 6 and 10. If the diameters are perpendicular, what is the area of the shaded regions?
     
    (A) 32π
    (B) 50π
    (C) 58π
    (D) 64π
    (E) 74π

STOP DOING MATH! You need almost none of it to solve this problem. How can you guesstimate this?

Well, what’s the area of the large circle? 100π. What’s half of that? 50π. Good, the answer is (B). Done.

Why? Glad you asked. What happens if I rearrange the pieces of the puzzle?

OH HELL YES. Look at that. Doesn’t that excite you? I love guesstimating so friggin’ much.

Note that guesstimating doesn’t just apply to shaded regions. You can use it to solve all kinds of geometry questions. As long as a figure is drawn to scale, you should ponder the implications of guesstimate for a few seconds before you start doing any math. This takes practice, but I promise you it’s worth it when you become proficient.

The questions below can be solved with math, but your mission is to solve them with guesstimate. Make me proud.

Don’t stop! Don’t ever stop!

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  1. In the figure above, AB is the diameter of the circle, and AC = BC. What is the area of the shaded region?
     
    (A) 4π – 2
    (B) 2π – 1
    (C) π
    (D) π – 1
    (E) π – 2

Answer and explanation after the jump…

As is usually the case with shaded region problems, the easiest way (and in this case, the only way) to solve for the shaded region is to solve for the regions around it first. The relationship we want to keep in mind is:

In this particular case, we’re going to have to do a little legwork to figure out what out “whole” is before we get down to business.

Let’s start by dropping a vertical from the top of our isosceles triangle (and noting that in doing so, we’re drawing a radius, so it’s got a length of 2):

That vertical is of course perpendicular to AB, and creates a right angle that nicely frames the area we’re looking to solve for. So the Areawhole we’re looking for here is actually only the part of the circle marked off by that right angle. Since a circle has 360 degrees of arc and we’re only dealing with 90 of them, we’re dealing with one fourth of the circle.

Areacircleπr2
Areacircle = π(22)
Areacircle = 4π

So the area of the sector we care about is simply one fourth of that, or π:

Areawholeπ

Now we just need to find the area of the unshaded part (the right triangle we created, in red):

Areaunshaded = 1/2 (2)(2)
Areaunshaded = 2

So the area of our shaded region must be…

Areashadedπ2

That’s answer choice (E). 

One more note about this one: since the diagram is drawn to scale, it’s possible (and wise) to use your guesstimation skills once you’ve found an answer (or even to eliminate answers before you do much calculating). Which is to say: when you know the area of the triangle is 2, does it make sense that the shaded region is about 1.14159? I’d say, by eyeballing it, that yeah, it does. It wouldn’t have made sense, though, had we made a math mistake somehow and ended up with a different choice, like (D) π – 1. To pick that answer would be to say (insanely) that the shaded region is as big as the right triangle we made by dropping that vertical. That’s crazy talk.